Difference between revisions of "1950 AHSME Problems/Problem 35"

m (Solution)
(Solution)
(One intermediate revision by one other user not shown)
Line 9: Line 9:
  
 
==Solution==
 
==Solution==
The inradius is equal to the area divided by semiperimeter. The area is <math>\frac{(10)(24)}{2} = 120</math> because it's a right triangle. The semiperimeter is <math>30</math>. Therefore the inradius is <math>\boxed{\textbf{(B)}\ 4}</math>.
+
The inradius is equal to the area divided by semiperimeter. The area is <math>\frac{(10)(24)}{2} = 120</math> because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is <math>30</math>. Therefore the inradius is <math>\boxed{\textbf{(B)}\ 4}</math>.
  
 
==See Also==
 
==See Also==
Line 15: Line 15:
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 14:43, 29 November 2019

Problem

In triangle $ABC$, $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:

$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$

Solution

The inradius is equal to the area divided by semiperimeter. The area is $\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$. Therefore the inradius is $\boxed{\textbf{(B)}\ 4}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 34
Followed by
Problem 36
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png